giải pt : a, √5x+6=x-6 b, |x-2|=2x-1 c, √2x-3=x-2 d, |x-4|-2x+2=0 19/11/2021 Bởi Katherine giải pt : a, √5x+6=x-6 b, |x-2|=2x-1 c, √2x-3=x-2 d, |x-4|-2x+2=0
Đáp án: d) x=2 Giải thích các bước giải: \(\begin{array}{l}a)\sqrt {5x + 6} = x – 6\\ \to 5x + 6 = {x^2} – 12x + 36\left( {DK:x \ge 6} \right)\\ \to {x^2} – 17x + 30 = 0\\ \to \left[ \begin{array}{l}x = 15\left( {TM} \right)\\x = 2\left( l \right)\end{array} \right.\\b)\left| {x – 2} \right| = 2x – 1\\ \to {x^2} – 4x + 4 = 4{x^2} – 4x + 1\left( {DK:x \ge \dfrac{1}{2}} \right)\\ \to 3{x^2} = 3\\ \to \left[ \begin{array}{l}x = 1\left( {TM} \right)\\x = – 1\left( l \right)\end{array} \right.\\c)\sqrt {2x – 3} = x – 2\\ \to 2x – 3 = {x^2} – 4x + 4\left( {DK:x \ge 2} \right)\\ \to {x^2} – 6x + 7 = 0\\ \to \left[ \begin{array}{l}x = 3 + \sqrt 2 \left( {TM} \right)\\x = 3 – \sqrt 2 \left( l \right)\end{array} \right.\\d)\left| {x – 4} \right| – 2x + 2 = 0\\ \to \left| {x – 4} \right| = 2x – 2\\ \to {x^2} – 8x + 16 = 4{x^2} – 8x + 4\left( {DK:x \ge 1} \right)\\ \to 3{x^2} = 12\\ \to {x^2} = 4\\ \to \left[ \begin{array}{l}x = 2\left( {TM} \right)\\x = – 2\left( l \right)\end{array} \right.\end{array}\) Bình luận
Giải thích các bước giải: a) ⇔$\left \{ {{x – 6 ≥ 0} \atop {5x+6=(x-6)^{2} }} \right.$ ⇔$\left \{ {{x≥-6} \atop {5x+6=x^2-12x+36}} \right.$ ⇔$\left \{ {{x≥-6} \atop {-x^2+17x-30=0}} \right.$ ⇔$\left \{ {{x≥-6} \atop {\left[ \begin{array}{l}x=15(TM)\\x=2(L)\end{array} \right.\ }} \right.$ c) ⇔ $\left \{ {{x-2≥0} \atop {2x-3=(x-2)^2}} \right.$ ⇔$\left \{ {{x≥2} \atop {2x-3=x^2-4x+4}} \right.$ ⇔$\left \{ {{x≥2} \atop {-x^2+6x-7=0}} \right.$ ⇔$\left \{ {{x≥2} \atop { \left[ \begin{array}{l}x=3+\sqrt{2}(TM) \\x=3-\sqrt{2}(L) \end{array} \right.\ }} \right.$ Bình luận
Đáp án:
d) x=2
Giải thích các bước giải:
\(\begin{array}{l}
a)\sqrt {5x + 6} = x – 6\\
\to 5x + 6 = {x^2} – 12x + 36\left( {DK:x \ge 6} \right)\\
\to {x^2} – 17x + 30 = 0\\
\to \left[ \begin{array}{l}
x = 15\left( {TM} \right)\\
x = 2\left( l \right)
\end{array} \right.\\
b)\left| {x – 2} \right| = 2x – 1\\
\to {x^2} – 4x + 4 = 4{x^2} – 4x + 1\left( {DK:x \ge \dfrac{1}{2}} \right)\\
\to 3{x^2} = 3\\
\to \left[ \begin{array}{l}
x = 1\left( {TM} \right)\\
x = – 1\left( l \right)
\end{array} \right.\\
c)\sqrt {2x – 3} = x – 2\\
\to 2x – 3 = {x^2} – 4x + 4\left( {DK:x \ge 2} \right)\\
\to {x^2} – 6x + 7 = 0\\
\to \left[ \begin{array}{l}
x = 3 + \sqrt 2 \left( {TM} \right)\\
x = 3 – \sqrt 2 \left( l \right)
\end{array} \right.\\
d)\left| {x – 4} \right| – 2x + 2 = 0\\
\to \left| {x – 4} \right| = 2x – 2\\
\to {x^2} – 8x + 16 = 4{x^2} – 8x + 4\left( {DK:x \ge 1} \right)\\
\to 3{x^2} = 12\\
\to {x^2} = 4\\
\to \left[ \begin{array}{l}
x = 2\left( {TM} \right)\\
x = – 2\left( l \right)
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
a)
⇔$\left \{ {{x – 6 ≥ 0} \atop {5x+6=(x-6)^{2} }} \right.$
⇔$\left \{ {{x≥-6} \atop {5x+6=x^2-12x+36}} \right.$
⇔$\left \{ {{x≥-6} \atop {-x^2+17x-30=0}} \right.$
⇔$\left \{ {{x≥-6} \atop {\left[ \begin{array}{l}x=15(TM)\\x=2(L)\end{array} \right.\ }} \right.$
c)
⇔ $\left \{ {{x-2≥0} \atop {2x-3=(x-2)^2}} \right.$
⇔$\left \{ {{x≥2} \atop {2x-3=x^2-4x+4}} \right.$
⇔$\left \{ {{x≥2} \atop {-x^2+6x-7=0}} \right.$
⇔$\left \{ {{x≥2} \atop { \left[ \begin{array}{l}x=3+\sqrt{2}(TM) \\x=3-\sqrt{2}(L) \end{array} \right.\ }} \right.$